Newton’s Backward Difference Interpolation

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2021)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

Newton’s backward interpolation formula is the natural companion to the forward formula: use backward differences when the interpolation point lies near the end of the data table. UPSC 2021 tested it as a Section B derivation-and-compute: construct the backward difference table and then apply the formula. Showing the difference table in full is essential for both the derivation marks and the computation marks.

Minimum Theory

Setup. Given $n+1$ equally-spaced data points

$x_0, x_1, \ldots, x_n \quad \text{with step } h = x_i - x_{i-1}$

and corresponding values $y_0, y_1, \ldots, y_n$.

Backward difference operator $\nabla$.

$\nabla y_i = y_i - y_{i-1}$

$\nabla^2 y_i = \nabla y_i - \nabla y_{i-1} = y_i - 2y_{i-1} + y_{i-2}$

$\nabla^k y_i = \nabla^{k-1} y_i - \nabla^{k-1} y_{i-1}$

Each column of the backward difference table is formed by subtracting the entry above-left from the current entry.

Newton’s backward interpolation formula.

Let $s = (x - x_n)/h$ (note: $s \le 0$ when $x \le x_n$).

$y(x) = y_n + s\,\nabla y_n + \frac{s(s+1)}{2!}\,\nabla^2 y_n + \frac{s(s+1)(s+2)}{3!}\,\nabla^3 y_n + \cdots$

In compact form:

$y(x) = \sum_{k=0}^{n} \binom{s+k-1}{k}(-1)^k(-1)^k \nabla^k y_n = \sum_{k=0}^{n} \frac{s(s+1)\cdots(s+k-1)}{k!}\,\nabla^k y_n$

where the $k=0$ term is $y_n$ (empty product equals 1).

When to use backward vs. forward.

Derivation sketch. The formula follows from writing the interpolating polynomial through $x_n, x_{n-1}, \ldots, x_0$ in Newton’s divided-difference form, then specialising to equal spacing. The backward differences $\nabla^k y_n$ replace the divided differences.

Truncation error. The error of the $n$-th degree interpolant at $x$ is

$E = \frac{s(s+1)\cdots(s+n)}{(n+1)!}\,h^{n+1}\,f^{(n+1)}(\xi)$

for some $\xi$ in the range of the data.

Question Archetypes

table-and-interpolate (1 question; 2021)

Recognition Cues

• A table of equally-spaced $x$ and $y$ values is provided.
• “Use Newton’s backward interpolation formula to find $y$ at $x = \cdots$.”
• The target $x$ is near the last entry of the table.

Solution Template

1. Write out the data table and identify $h$ and $x_n$.
2. Construct the full backward difference table column by column.
3. Read off $y_n, \nabla y_n, \nabla^2 y_n, \ldots$ from the last row (the row for $x_n$).
4. Compute $s = (x - x_n)/h$.
5. Substitute into the formula; compute each term and sum.
6. State the interpolated value.

Worked Example

2021 Paper 2, 2021-P2-Q3b (15 marks)

The following table gives values of a function $f(x)$:

$x$	1.0	1.2	1.4	1.6	1.8
$y$	2.7183	3.3201	4.0552	4.9530	6.0496

Using Newton’s backward interpolation formula, estimate $f(1.75)$.

Step 1 — identify parameters.

$h = 0.2$, $x_n = x_4 = 1.8$, $y_n = y_4 = 6.0496$.

Step 2 — backward difference table.

Construct column by column; each entry = current $y$ minus the entry one row up in the previous column.

Values used from the last row (bottom of each column):

$y_4 = 6.0496,\quad \nabla y_4 = 1.0966,\quad \nabla^2 y_4 = 0.1988,\quad \nabla^3 y_4 = 0.0361,\quad \nabla^4 y_4 = 0.0067$

Step 3 — compute $s$.

$s = \frac{x - x_n}{h} = \frac{1.75 - 1.8}{0.2} = \frac{-0.05}{0.2} = -0.25$

Step 4 — apply the formula.

$y(1.75) = y_4 + s\,\nabla y_4 + \frac{s(s+1)}{2!}\,\nabla^2 y_4 + \frac{s(s+1)(s+2)}{3!}\,\nabla^3 y_4 + \frac{s(s+1)(s+2)(s+3)}{4!}\,\nabla^4 y_4$

Term 0: $6.0496$

Term 1: $(-0.25)(1.0966) = -0.27415$

Term 2: $\dfrac{(-0.25)(0.75)}{2}(0.1988) = \dfrac{-0.1875}{2}(0.1988) = (-0.09375)(0.1988) = -0.018638$

Term 3: $\dfrac{(-0.25)(0.75)(1.75)}{6}(0.0361) = \dfrac{-0.328125}{6}(0.0361) = (-0.054688)(0.0361) = -0.001974$

Term 4: $\dfrac{(-0.25)(0.75)(1.75)(2.75)}{24}(0.0067) = \dfrac{-0.902344}{24}(0.0067) = (-0.037598)(0.0067) = -0.000252$

Step 5 — sum.

$y(1.75) \approx 6.0496 - 0.27415 - 0.01864 - 0.00197 - 0.00025 = 5.75459$

$\boxed{f(1.75) \approx 5.7546}$

(The true value $e^{1.75} \approx 5.7546$ confirms the result.)

Common Traps

• Using $s = (x - x_0)/h$ (that is the forward formula’s $s$). For backward interpolation, $s = (x - x_n)/h$, which is negative when $x < x_n$.
• Forming the difference table by subtracting in the wrong direction: each backward difference is “current minus previous” ($y_i - y_{i-1}$), not the other way.
• Reading the wrong diagonal: all values used in the formula come from the last row of the backward difference table, not from the first row or the main diagonal.
• Dropping the $(s+k-1)$ pattern in the numerators — the terms are $s$, $s(s+1)$, $s(s+1)(s+2)$, …; each new factor adds $+1$ to the last factor, not $-1$.

Marks-Aware Writing

At 15 marks in Section B, the backward difference table alone is worth 5–6 marks. Write it as a clear array with column headers $\nabla y$, $\nabla^2 y$, etc. Then devote a line to computing $s$, and write out each term of the formula separately before summing — this is the cleanest way to show work and collect all method marks. A bare final answer will lose most marks even if numerically correct.

Practice Set

Only one historical question on this atom (shown above).